What is the integral from 1 to 2 of (6xy^2-4 x/y) ?

The integral from 1 to 2 of (6xy^2-4 x/y) is -4xln(2)+14x

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integral from 1 to 2 of (6xy^2-4 x/y)

\int_{1}^{2} (6 x y^{2} - 4 \frac{x}{y}) d y

- 4 x ln (2) + 14 x

Solution steps

\int_{1}^{2} 6 x y^{2} - 4 \cdot \frac{x}{y} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{1}^{2} 4 \cdot \frac{x}{y} d y + \int_{1}^{2} 6 x y^{2} d y

\int_{1}^{2} 4 \cdot \frac{x}{y} d y = 4 x ln (2)

\int_{1}^{2} 6 x y^{2} d y = 14 x

= - 4 x ln (2) + 14 x

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What is the integral from 1 to 2 of (6xy^2-4 x/y) ?
The integral from 1 to 2 of (6xy^2-4 x/y) is -4xln(2)+14x